1. Answer the following a) Determine the number of electrons transferred for 1.0

Question

1. Answer the following

a) Determine the number of electrons transferred for 1.0 g of Fe (s) reacted.

Consider the following overall reaction for the formation of rust, iron(III) oxide.

Fe (s) + O2 (g) Fe2O3 (s)

b) Calculate G° (in kJ) for the following redox reaction:

Cu2+ (aq) +   Zn (s)    Cu (s) +   Zn2+ (aq)

Given:
Zn2+(aq) + 2 e– Zn(s)      Eo= -0.76 V
Cu2+(aq) + 2 e– Cu(s)       Eo= 0.34 V
Use F = 96485 C/V.mol

c)

A voltaic cell was constructed using the half-cell Cr(s) | Cr3+(0.20 M) as the anode and the half-cell Sn2+(x M) | Sn(s) as the cathode; the measured cell voltage was 0.56 V. Calculate x, the molarity of Sn2+ in the cathode.

Cr3+(aq) + 3 e– Cr(s)    Eo = -0.74 V

Sn2+(aq) + 2 e– Sn(s)    Eo = -0.14 V

Explanation / Answer

a)

Fe (s) + O2 (g) Fe2O3 (s)

Fe goes from 0 to Fe+3

O goes from 0 to -2 so

3*2 = 6 electrons are being transferred

MW of Fe = 55.8450

mol = mass/MW = 1/55.8450 = 0.017906 mol of Fe

each mol of Fe --> 3 mol of e-

so

0.017906 mol o FE = 3*0.017906 = 0.053718 mol of e- are trasferred

1 mol = 6.022*10^23

then

0.053718 mol --> 0.053718* 6.022*10^23 =3.2348*10^22 electrons

b)

E°cell = Ered - Eox

Ered = reduction, Eox = oxidation

the larges potential -> 0.34 so

Cu is being reduced

E°cell = 0.34 -- 0.76 = 1.1 V

s

dG = -nFE°cell

dG = -RT*ln(K)

so

RTln(K) = nF*E°Cell

K =exp( (nF)/(RT) * E°cell)

K =exp( (2*96500)/(8.314*298) *1.1) = 1.637*10^37

c)

find

Ecell = E°cell - 00592/n*log(Q)

Q = [Cr+3]^/[Sn+2]^3

E°cell = -0.14 - -0.74 = 0.6 V

substitute

0.56 = 0.60 -0.0592/6 * log(0.20^2)/(X^3))

10^((0.56 - 0.60 )*6/(-0.0592)) = 0.20^2)/(X^3)

11325.30 = (0.20^2)/(X^3)

x^3 = (0.2^2) / (11325.30 )

x = 0.00000353191^(1/3) = 0.015228M for SN+2

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.